In a $\Delta ABC$, if $r_1=1$ and $r_2=2$ and $C=\frac{\pi}{2}$, find area of triangle.

Question

$r_1$ and $r_2$ are ex-radii about side a and b respectively. $\Delta$ is area of triangle. s is the semi-perimeter

Since $r_1=\frac{\Delta}{s-a}$ $$\Delta=s-a$$ and $$\Delta=2(s-b)$$

From this, it can be observed that $$b-a=\frac{\Delta}{2}$$ and $$c=\frac{3\Delta}{2}$$

How should I proceed ?

Answer 1

Using $\Delta = (S-a)=2(S-b)$ Now replace $S=\frac{a+b+c}{2}$ we get $$ c=3(b-a) --(1)\,and\,a^2+b^2=c^2--(2)$$ Also using $r_1=\frac{\Delta}{S-a}=1$ $$\frac{1}{2}ab=\frac{b+c-a}{2}$$ $$ab=\frac{4}{3}c-----(3)$$ Squaring equation (1) $$9(a^2+b^2-2ab)=c^2$$ Using (3) we get $$9c^2-24c=c^2$$ Hence $c=3$ using equation (3) $\Delta =2$

Answer 2

\begin{align} r_a\,r_b&=\frac{S_{ABC}^2}{(\rho-a)(\rho-b)} =\rho\,(\rho-c) =\frac{S_{ABC}}{\tan\tfrac C2} ,\\ S_{ABC}&=r_a\,r_b\,\tan\tfrac C2 =2 . \end{align}